Blog: `x^3/((x + 2)^2(x - 2)^2)` Write the partial fraction decomposition of the rational expression.

Monday, August 17, 2009

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$

Let $\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{A}}{{{x}+{2}}}+\frac{{B}}{{{\left({x}+{2}\right)}}^{{2}}}+\frac{{C}}{{{x}-{2}}}+\frac{{D}}{{{\left({x}-{2}\right)}}^{{2}}}$

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{{A}{\left({x}+{2}\right)}{{\left({x}-{2}\right)}}^{{2}}+{B}{{\left({x}-{2}\right)}}^{{2}}+{C}{\left({x}-{2}\right)}{{\left({x}+{2}\right)}}^{{2}}+{D}{{\left({x}+{2}\right)}}^{{2}}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{{A}{\left({x}+{2}\right)}{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{B}{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{C}{\left({x}-{2}\right)}{\left({{x}}^{{2}}+{4}{x}+{4}\right)}+{D}{\left({{x}}^{{2}}+{4}{x}+{4}\right)}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{{A}{\left({{x}}^{{3}}-{4}{{x}}^{{2}}+{4}{x}+{2}{{x}}^{{2}}-{8}{x}+{8}\right)}+{B}{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{C}{\left({{x}}^{{3}}+{4}{{x}}^{{2}}+{4}{x}-{2}{{x}}^{{2}}-{8}{x}-{8}\right)}+{D}{\left({{x}}^{{2}}+{4}{x}+{4}\right)}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{{A}{\left({{x}}^{{3}}-{2}{{x}}^{{2}}-{4}{x}+{8}\right)}+{B}{\left({{x}}^{{2}}-{4}{x}+{4}\right)}+{C}{\left({{x}}^{{3}}+{2}{{x}}^{{2}}-{4}{x}-{8}\right)}+{D}{\left({{x}}^{{2}}+{4}{x}+{4}\right)}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{{{x}}^{{3}}{\left({A}+{C}\right)}+{{x}}^{{2}}{\left(-{2}{A}+{B}+{2}{C}+{D}\right)}+{x}{\left(-{4}{A}-{4}{B}-{4}{C}+{4}{D}\right)}+{8}{A}+{4}{B}-{8}{C}+{4}{D}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$

$\displaystyle\therefore{{x}}^{{3}}={{x}}^{{3}}{\left({A}+{C}\right)}+{{x}}^{{2}}{\left(-{2}{A}+{B}+{2}{C}+{D}\right)}+{x}{\left(-{4}{A}-{4}{B}-{4}{C}+{4}{D}\right)}+{8}{A}+{4}{B}-{8}{C}+{4}{D}$

equating the coefficients of the like terms,

$\displaystyle{A}+{C}={1}$

$\displaystyle-{2}{A}+{B}+{2}{C}+{D}={0}$

$\displaystyle-{4}{A}-{4}{B}-{4}{C}+{4}{D}={0}$

$\displaystyle{8}{A}+{4}{B}-{8}{C}+{4}{D}={0}$

Now we have to solve the above four equations to get the solutions of A,B,C and D.

Rewrite the third equation as,

$\displaystyle-{4}{\left({A}+{C}\right)}-{4}{B}+{4}{D}={0}$

substitute the expression of ( A+C) from the first equation in the above equation,

$\displaystyle-{4}{\left({1}\right)}-{4}{B}+{4}{D}={0}$

$\displaystyle-{4}{B}+{4}{D}={4}$

$\displaystyle{4}{\left(-{B}+{D}\right)}={4}$

$\displaystyle-{B}+{D}={1}$

$\displaystyle{D}={1}+{B}$

Express C in terms of A from first equation,

$\displaystyle{C}={1}-{A}$

Substitute the expressions of C and D in the second equation,

$\displaystyle-{2}{A}+{B}+{2}{\left({1}-{A}\right)}+{1}+{B}={0}$

$\displaystyle-{2}{A}+{B}+{2}-{2}{A}+{1}+{B}={0}$

$\displaystyle-{4}{A}+{2}{B}=-{3}$ (equation 5)

Substitute the expressions of C and D in the fourth equation,

$\displaystyle{8}{A}+{4}{B}-{8}{\left({1}-{A}\right)}+{4}{\left({1}+{B}\right)}={0}$

$\displaystyle{8}{A}+{4}{B}-{8}+{8}{A}+{4}+{4}{B}={0}$

$\displaystyle{16}{A}+{8}{B}={4}$

$\displaystyle{4}{\left({4}{A}+{2}{B}\right)}={4}$

$\displaystyle{4}{A}+{2}{B}={1}$ (equation 6)

Solve equation 5 and 6 to get the solutions of A and B,

Add equation 5 and 6,

$\displaystyle{4}{B}=-{3}+{1}$

$\displaystyle{4}{B}=-{2}$

$\displaystyle{B}=-\frac{{2}}{{4}}$

$\displaystyle{B}=-\frac{{1}}{{2}}$

Plug the value of B in the equation 5,

$\displaystyle-{4}{A}+{2}{\left(-\frac{{1}}{{2}}\right)}=-{3}$

$\displaystyle-{4}{A}-{1}=-{3}$

$\displaystyle-{4}{A}=-{3}+{1}$

$\displaystyle-{4}{A}=-{2}$

$\displaystyle{A}=\frac{{1}}{{2}}$

Now plug the values of A and B in the expressions of C and D,

$\displaystyle{C}={1}-\frac{{1}}{{2}}$

$\displaystyle{C}=\frac{{1}}{{2}}$

$\displaystyle{D}={1}+{\left(-\frac{{1}}{{2}}\right)}$

$\displaystyle{D}=\frac{{1}}{{2}}$

$\displaystyle\therefore\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}=\frac{{1}}{{{2}{\left({x}+{2}\right)}}}-\frac{{1}}{{{2}{{\left({x}+{2}\right)}}^{{2}}}}+\frac{{1}}{{{2}{\left({x}-{2}\right)}}}+\frac{{1}}{{{2}{{\left({x}-{2}\right)}}^{{2}}}}$

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Monday, August 17, 2009

$\displaystyle\frac{{{x}}^{{3}}}{{{{\left({x}+{2}\right)}}^{{2}}{{\left({x}-{2}\right)}}^{{2}}}}$ Write the partial fraction decomposition of the rational expression.

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